Archive for February, 2017

For some time now we’ve been analyzing the helpfulness of the engineering phenomena known as pulleys and we’ve learned that, yes, they can be very helpful, although they do have their limitations. One of those ever-present limitations is due to the inevitable presence of friction between moving parts. Like an unsummoned gremlin, friction will be standing by in any mechanical situation to put the wrench in the works. Today we’ll calculate just how much friction is present within the examplecompound pulley we’ve been working with.

So How Much Friction is Present in our Compound Pulley?

Last time we began our numerical demonstration of the inequality between a compound pulley’swork input, WI, and work output, WO, an inequality that’s due to friction in its wheels. We began things by examining a friction-free scenario and discovered that to lift an urn with a weight, W, of 40 pounds a distance, d1, of 2 feet above the ground, Mr. Toga exerts a personal effort/force, F, of 10 pounds to extract a length of rope, d2, of 8 feet.

In reality our compound pulley must contend with the effects of friction, so we know it will take more than 10 pounds of force to lift the urn, a resistance which we’ll notate FF. To determine this value we’ll attach a spring scale to Mr. Toga’s end of the rope and measure his actual lifting force, FActual, represented by the formula,

FActual =F + FF (1)

We find that FActual equals 12 pounds. Thus our equation becomes,

12 Lbs = 10 Lbs + FF (2)

which simplifies to,

2 Lbs =FF (3)

Now that we’ve determined values for all operating variables, we can solve for work input and then contrast our finding with work output,

It’s evident that the amount of work Mr. Toga puts into lifting his urn requires 16 more Foot-Pounds of work input effort than the amount of work output produced. This extra effort that’s required to overcome the pulley’s friction is the same as the work required to carry a weight of one pound a distance of 16 feet. We can thus conclude that work input does not equal work output in a compound pulley.

Next time we’ll take a look at a different use for pulleys beyond that of just lifting objects.

Last time we began work on a numericaldemonstration and engineering analysis of the inequality of work input and output as experienced by our example persona, an ancient Greek lifting an urn. Today we’ll get two steps closer to demonstrating this reality as we work a compound pulley’s numerical puzzle, shuffling equations like a Rubik’s Cube to arrive at values for two variables crucial to our analysis, d2, the length of rope he extracts from the pulley while lifting, and F, the force/effort required to lift the urn in an idealized situation where no friction exists.

A Compound Pulley’s Numerical Puzzle is Like a Rubik’s Cube

We’ll continue manipulating the work input equation, WI, as shown in Equation (1), along with derivative equations, breaking it down into parts, and handle the two terms within parentheses separately. Term one, (F × d2), corresponds to the force/effort/work required to lift the urn in an idealized no-friction world. It’ll be our focus today as it provides a springboard to solving for variables F and d2.

WI = (F ×d2) + (FF×d2) (1)

Previously we learned that when friction is present, work output, WO, is equal to work input minus the work required to oppose friction while lifting. Mathematically that’s represented by,

WO = WI – (FF × d2) (2)

We also previously calculated WO to equal 80 Ft-Lbs. To get F and d2 into a relationship with terms we already know the value for, namely WO, we substitute Equation (1) into Equation (2) and arrive at,

And because in our example four ropes are used to support the weight of the urn, we know that MA equals 4. We also know from last time that d1 equals 2 feet. Plugging these numbers into Equation (5) we arrive at a value for d2,

d2 ÷ 2 ft= 4 (6)

d2= 4 × 2 ft (7)

d2 = 8 ft (8)

Substituting Equation (8) into Equation (4), we solve for F,

80 Ft-Lbs =F × 8 ft (9)

F = 10 Pounds (10)

Now that we know F and d2 we can solve for FF, the amount of extra effort required by man or machine to overcome friction in a compound pulley assembly. It’s the final piece in the numerical puzzle which will then allow us to compare work input to output.

Last time we performed an engineering analysis of a compound pulley which resulted in an equation comparing the amount of true work effort, or work input,WI, required by machine or human to lift an object, in our case a toga’d man lifting an urn. Our analysis revealed that, in real world situations, work input does not equal work output, WO, due to the presence of friction. Today we’ll begin to numerically demonstrate their inequality by first solving for work output, and later work input.

In our example Mr. Toga lifts an urn of weight, W, equal to 40 pounds to a height, or distance off the floor, d1, of 2 feet. Inserting these values into equation (1) we arrive at,

WO = 40 pounds ×2 feet= 80 Ft-Lbs (2)

where, Ft-Lbs is a unit of work which denotes pounds of force moving through feet of distance.

Now that we’ve calculated thework output, we’ll turn our attention to the previously-derived equation for work input, shown in equation (3). Interrelating equations for WO and WI will enable us to solve for unknown variables, including the force, F, required to lift the urn and the length of rope, d2, extracted during lifting.Once F and d2 are known, we can solve for the additional force required to overcome friction, FF, then finally we’ll solve for WI.

Once again, the equation we’ll be working with is,

WI = (F ×d2) + (FF×d2) (3)

To calculate F, we’ll work the two terms present within parentheses separately, then use knowledge gained to further work our way towards a numerical comparison of work input and work output. We’ll do that next time.